Q107 / 200 · Manipal (MET) 2013 · Mathematics
If $\sin ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$, then $\frac{d^2 y}{d x^2}$ equals
- A$\frac{x}{y}$
- B$\frac{y}{x^2}$
- C$\frac{y}{x}$
- Correct answerD$0$
Solution
We have $\sin ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=\log a$
$\Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\sin (\log a)$
$\Rightarrow \quad \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\sin (\log a)$
(on putting $y=x \tan \theta$.)
$\begin{aligned} \Rightarrow & & \cos 2 \theta & =\sin (\log a) \\ \Rightarrow & & 2 \theta & =\cos ^{-1}(\sin \log a) \\ \Rightarrow & & \theta & =\frac{1}{2} \cos ^{-1}\{\sin (\log a)\}\end{aligned}$
$\tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \cos ^{-1}\{\sin \log a\}$
$\Rightarrow \frac{1}{1+\frac{y^2}{x^2}} \cdot \frac{x \frac{d y}{d x}-y}{x^2}=0$
$\Rightarrow \quad x \frac{d y}{d x}-y=0$
$\Rightarrow \quad \frac{d y}{d x}=\frac{y}{x}$ ....(i)
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-y}{x^2}+\frac{1}{x} \cdot \frac{d y}{d x}=\frac{-y}{x^2}+\frac{1}{x}\left(\frac{y}{x}\right) ; \quad\{$ from $(i)\}$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-y}{x^2}+\frac{y}{x^2}=0$.
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-y}{x^2}+\frac{y}{x^2}=0$.
Hence, the correct answer is option: 'D'.